In an equilateral triangle ABC, D is a point on the side BC
In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD2 = 7AB2
Solution
Construction: Draw AP ⟂ BC
In Δ ADP,
AD2 = AP2 + DP2
AD2 = AP2 + (BP - BD)2
AD2 = AP2 + BP2 + BD2 - 2(BP)(BD)
AD2 = AB2 + (1/3 BC)2 - 2(BC/2)(BC/3)
AD2 = 7/9 AB2 (Since BC = AB)
9AD2 = 7AB2